Let $V$ be the elliptic curve V: $x^3$+ $y^3$ = A$z^3$ where A > 2 is cube free natural number. A conjugate quadratic point of $V$ is one of the form $(a + b\sqrt d, a - b\sqrt d, c)$ (note that all quadratic point $(x, y, z)$ of $V$ can be reduced to have $z$ rational integer multiplying by the conjugate of $z$). I have found the result which follows and I am interested in knowing the opinion and remarks or possible objections of readers and mainly I expect for another proof.
Prove that $V(\mathbb Q)$ has no rational points distinct of $(1, -1, 0)$ if and only if all the quadratic points of $V$ are conjugates.
EXAMPLE.- This is true even with $x^3$ + $y^3$ = 2$z^3$ in which the only rational points are $(1, 1, 1)$ and $(1, -1, 0)$. One has here, with $A = 2$, the point $(4 +2\sqrt{-11}, -1 + \sqrt{-11}, -6)$ which is not conjugate because of the rational point $(1, 1, 1)$.
